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Differential Calculus Unit 6 Review

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6.1 Product rule

6.1 Product rule

Written by the Fiveable Content Team • Last updated August 2025
Written by the Fiveable Content Team • Last updated August 2025
Differential Calculus
Unit & Topic Study Guides

Product, Quotient Rules & Higher-Order Derivatives

The product rule is a key tool for differentiating functions that are multiplied together. It breaks down complex expressions into simpler parts, making it easier to find derivatives of tricky functions.

Using this rule, you can tackle derivatives of products involving polynomials, trig functions, and exponentials. It's essential to know when to use it and how to combine it with other differentiation rules for more complex problems.

The Product Rule

Product rule for differentiation

  • States that the derivative of the product of two differentiable functions f(x)f(x) and g(x)g(x) equals:
    • f(x)f(x) times the derivative of g(x)g(x), plus
    • g(x)g(x) times the derivative of f(x)f(x)
  • Formally written as: ddx[f(x)g(x)]=f(x)ddx[g(x)]+g(x)ddx[f(x)]\frac{d}{dx}[f(x)g(x)] = f(x)\frac{d}{dx}[g(x)] + g(x)\frac{d}{dx}[f(x)]
  • Useful for finding derivatives of products of functions (polynomials, trigonometric functions, exponentials)
  • To apply the rule:
    1. Identify the two functions being multiplied
    2. Find the derivative of each function separately
    3. Multiply the first function by the derivative of the second
    4. Multiply the second function by the derivative of the first
    5. Add the resulting terms together
Product rule for differentiation, High School Calculus/Techniques of Differentiation - Wikibooks, open books for an open world

Application of product rule

  • Example 1: Derivative of f(x)=x2sin(x)f(x) = x^2 \sin(x)
    • Let u=x2u = x^2 and v=sin(x)v = \sin(x)
    • ddx[u]=2x\frac{d}{dx}[u] = 2x and ddx[v]=cos(x)\frac{d}{dx}[v] = \cos(x)
    • Applying the product rule: ddx[f(x)]=x2cos(x)+2xsin(x)\frac{d}{dx}[f(x)] = x^2 \cos(x) + 2x \sin(x)
  • Example 2: Derivative of g(x)=(3x+1)(x32x)g(x) = (3x + 1)(x^3 - 2x)
    • Let u=3x+1u = 3x + 1 and v=x32xv = x^3 - 2x
    • ddx[u]=3\frac{d}{dx}[u] = 3 and ddx[v]=3x22\frac{d}{dx}[v] = 3x^2 - 2
    • Applying the product rule: ddx[g(x)]=(3x+1)(3x22)+(x32x)(3)\frac{d}{dx}[g(x)] = (3x + 1)(3x^2 - 2) + (x^3 - 2x)(3)
  • Can be applied repeatedly for products of more than two functions
Product rule for differentiation, Category:Product rule of differential calculus - Wikimedia Commons

Product rule vs other rules

  • Use the product rule when:
    • The function is a product of two or more differentiable functions
    • The functions being multiplied are not composed of one another
  • Do not use the product rule when:
    • The function is a sum or difference of terms (use sum rule)
    • The function is a quotient of two functions (use quotient rule)
    • The function is a composition of functions (use chain rule)
    • The function is a single term (use basic differentiation rules)

Combining product rule with other rules

  • Break complex expressions into simpler components
    • Identify products, quotients, compositions, sums/differences
    • Apply appropriate rule to each component
    • Combine results using rules for sums, differences, constant multiples
  • Example: Derivative of h(x)=(2x+1)(x23)3h(x) = (2x + 1)(x^2 - 3)^3
    • Let u=2x+1u = 2x + 1 and v=(x23)3v = (x^2 - 3)^3
    • ddx[u]=2\frac{d}{dx}[u] = 2 (basic differentiation rule)
    • To find ddx[v]\frac{d}{dx}[v], use chain rule:
      • Let w=x23w = x^2 - 3, then v=w3v = w^3
      • ddx[w]=2x\frac{d}{dx}[w] = 2x (basic differentiation rule)
      • ddx[v]=3w2ddx[w]=3(x23)2(2x)\frac{d}{dx}[v] = 3w^2 \frac{d}{dx}[w] = 3(x^2 - 3)^2(2x)
    • Apply product rule: ddx[h(x)]=(2x+1)(3(x23)2(2x))+((x23)3)(2)\frac{d}{dx}[h(x)] = (2x + 1)(3(x^2 - 3)^2(2x)) + ((x^2 - 3)^3)(2)
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